JEE MAIN Continuity Previous Year Questions (PYQs) – Page 2 of 4

Let the function $ f(x) = \begin{cases} \dfrac{\log_e(1+5x) - \log_e(1+\alpha x)}{x}, & x \neq 0 \\ 10, & x = 0 \end{cases} $ be continuous at $x=0$. Then $\alpha$ is equal to:

If the function $f$ defined as $f(x)=\dfrac{1}{x}-\dfrac{kx-1}{e^{2x}-1},\ x\ne0$, is continuous at $x=0$, then the ordered pair $(k,f(0))$ is equal to :

Let f : R $ \to $ R be defined as $f(x) = \left\{ \matrix{ 2\sin \left( { - {{\pi x} \over 2}} \right),if\,x < - 1 \hfill \cr |a{x^2} + x + b|,\,if - 1 \le x \le 1 \hfill \cr \sin (\pi x),\,if\,x > 1 \hfill \cr} \right.$ If f(x) is continuous on R, then a + b equals :

If a function $f(x)$ defined by  $f(x) = \begin{cases} ae^x + be^{-x}, & -1 \leq x < 1 \\[6pt] cx^2, & 1 \leq x \leq 3 \\[6pt] ax^2 + 2cx, & 3 < x \leq 4 \end{cases} \\[10pt] $ be continuous for some $ a, b, c \in \mathbb{R} $ and $f'(0) + f'(2) = e,$  then the value of $a$ is

Let $\alpha$ $\in$ R be such that the function $f(x) = \left\{ {\matrix{ {{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.$ is continuous at x = 0, where {x} = x $-$ [ x ] is the greatest integer less than or equal to x. Then :

$\mathrm{a}, \mathrm{b}>0$, let $f(x)= \begin{cases}\frac{\tan ((\mathrm{a}+1) x)+\mathrm{b} \tan x}{x}, & x< 0 \\ 3, & x=0 \\ \frac{\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2}-\sqrt{\mathrm{a} x}}{\mathrm{~b} \sqrt{\mathrm{a}} x \sqrt{x}}, & x> 0\end{cases}$ be a continuous function at $x=0$. Then $\frac{\mathrm{b}}{\mathrm{a}}$ is equal to :

If the function $f(x) = \left\{ {\matrix{ {(1 + |\cos x|)^{\lambda \over {|\cos x|}}} & , & {0 < x < {\pi \over 2}} \cr \mu & , & {x = {\pi \over 2}} \cr e^{{{\cot 6x} \over {{}\cot 4x}}} & , & {{\pi \over 2} < x < \pi } \cr } } \right.$

is continuous at $x = {\pi \over 2}$, then $9\lambda + 6{\log _e}\mu + {\mu ^6} - {e^{6\lambda }}$ is equal to

If the function $f$ defined on $\left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)$ by $f(x) = \begin{cases} \dfrac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x \ne \dfrac{\pi}{4} \ k, & x = \dfrac{\pi}{4} \end{cases}$ is continuous, then $k$ is equal to

Let f, g : R $\to$ R be functions defined by

$f(x) = \left\{ {\matrix{ {[x]} & , & {x < 0} \cr {|1 - x|} & , & {x \ge 0} \cr } } \right.$ and $g(x) = \left\{ {\matrix{ {{e^x} - x} & , & {x < 0} \cr {{{(x - 1)}^2} - 1} & , & {x \ge 0} \cr } } \right.$ where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly :

A ray of light coming from the point $P(1,2)$ gets reflected from the point $Q$ on the $x$-axis and then passes through the point $R(4,3)$. If the point $S(h,k)$ is such that $PQRS$ is a parallelogram, then $hk^{2}$ is equal to: